Uniform variant of Stirling's approximation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:18:16Z http://mathoverflow.net/feeds/question/28261 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28261/uniform-variant-of-stirlings-approximation Uniform variant of Stirling's approximation Matt Young 2010-06-15T15:11:40Z 2010-07-28T15:11:08Z <p>Stirling's formula is usually stated in the form $\log \Gamma(s) = (s-\frac12) \log{s} - s + \log\sqrt{2\pi} + E(s)$, where $E(s) = c_1/s + c_2/s^2 + \dots + O(s^{-K})$ for certain absolute constants $c_i$. I am interested in having a uniform approximation for $E(s)$ that is valid for all $s = \sigma + it$ with $\sigma>0$ fixed and $|t| \leq X$ for $X \geq 1$. Does there exist a known "nice" approximation for $E(s)$ of the form $E(s) = F(s) + O(X^{-K})$, where $F(s)$, which depends on $K$ and $X$ of course, has an explicit shape? Bonus points for explicit bounds<br> on $F(s)$ and its derivatives uniformly valid in $|t| \leq X$.</p> <p>EDIT ADDED July 28 2010: I am doubtful if there is a positive answer to my question. As a simple example, consider the rate of convergence of the Taylor series of the cosine function. Of course, $\cos(x) = 1- \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \pm \frac{x^{2k}}{(2k)!} + R_{2k}(x)$ where $R_{2k}(x) = \cos^{(2k+1)}(\xi) \frac{x^{2k+1}}{(2k+1)!}$ for some $|\xi| \leq |x|$. In order to get an error term that is $O(X^{-K})$ uniformly for $|x| \leq X$ we need to take $k$ roughly on the order of $X$ (since that is when the factorial in the denominator wins over the size of $X^{2k+1}$); at this point the error term gets very small very fast . This is a lot of terms!</p> http://mathoverflow.net/questions/28261/uniform-variant-of-stirlings-approximation/28321#28321 Answer by Fredrik Johansson for Uniform variant of Stirling's approximation Fredrik Johansson 2010-06-15T22:54:37Z 2010-06-15T22:54:37Z <p>Would the approximations of <a href="http://en.wikipedia.org/wiki/Lanczos_approximation" rel="nofollow">Lanczos</a> and <a href="http://en.wikipedia.org/wiki/Spouge%27s_approximation" rel="nofollow">Spouge</a> work? Spouge's approximation, in particular, has a rather nice form and an explicit error bound valid in the entire right half plane. Note that both approximations are for the gamma function, not the log gamma function, so getting the right branch might require additional work.</p> http://mathoverflow.net/questions/28261/uniform-variant-of-stirlings-approximation/33609#33609 Answer by J. M. for Uniform variant of Stirling's approximation J. M. 2010-07-28T03:43:36Z 2010-07-28T03:43:36Z <p>I would also recommend the Lanczos approximation. The article on Wikipedia links to <a href="http://rskey.org/gamma.htm" rel="nofollow">this article</a> which explains how the coefficients are generated, and gives an error bound on the approximation (this I think is more nicely formatted than the <a href="http://my.fit.edu/~gabdo/gamma.txt" rel="nofollow">original note</a> by Paul Godfrey, which also explains why the Spouge approximation might not be as good as Lanczos's).</p>